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(D) Let $C = \frac{\varepsilon_0 A}{d}$ be the capacitance of each pair of adjacent plates.From the figure, plates $1$ and $4$ are connected together.

Vedclass · Accepted Answer

$\frac{3}{2}\frac{{\varepsilon _0}A}{d}$

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(D) Let $C = \frac{\varepsilon_0 A}{d}$ be the capacitance of each pair of adjacent plates.From the figure, plates $1$ and $4$ are connected together. Plates $2$ and $3$ are connected to terminals $a$ and $b$ respectively.There are three capacitors formed by adjacent plates: $C_1$ (between $1$ and $2$), $C_2$ (between $2$ and $3$), and $C_3$ (between $3$ and $4$).Since plates $1$ and $4$ are connected, they are at the same potential. Let this potential be $V_0$.Terminal $a$ is connected to plate $2$ and terminal $b$ is connected to plate $3$.The capacitor $C_1$ is between plate $1$ (at $V_0$) and plate $2$ (at $V_a$).The capacitor $C_2$ is between plate $2$ (at $V_a$) and plate $3$ (at $V_b$).The capacitor $C_3$ is between plate $3$ (at $V_b$) and plate $4$ (at $V_0$).This forms a circuit where $C_1$ and $C_3$ are in parallel, and this combination is in series with $C_2$. However, looking at the provided circuit diagram, it shows $C_2$ in parallel with the series combination of $C_1$ and $C_3$.Equivalent capacitance $C_{eq} = C_2 + (C_1 	ext{ in series with } C_3) = C + \frac{C 	imes C}{C + C} = C + \frac{C}{2} = \frac{3}{2}C$.Substituting $C = \frac{\varepsilon_0 A}{d}$, we get $C_{ab} = \frac{3}{2} \frac{\varepsilon_0 A}{d}$.

Four metallic plates, each with a surface area of one side $A$, are placed at a distance $d$ from each other. The plates are connected as shown in the figure. The capacitance of the system between $a$ and $b$ is

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